US5742530A - Compact microelectronic device for performing modular multiplication and exponentiation over large numbers - Google Patents
Compact microelectronic device for performing modular multiplication and exponentiation over large numbers Download PDFInfo
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- US5742530A US5742530A US08/579,951 US57995195A US5742530A US 5742530 A US5742530 A US 5742530A US 57995195 A US57995195 A US 57995195A US 5742530 A US5742530 A US 5742530A
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- G—PHYSICS
- G06—COMPUTING; CALCULATING OR COUNTING
- G06F—ELECTRIC DIGITAL DATA PROCESSING
- G06F7/00—Methods or arrangements for processing data by operating upon the order or content of the data handled
- G06F7/60—Methods or arrangements for performing computations using a digital non-denominational number representation, i.e. number representation without radix; Computing devices using combinations of denominational and non-denominational quantity representations, e.g. using difunction pulse trains, STEELE computers, phase computers
- G06F7/72—Methods or arrangements for performing computations using a digital non-denominational number representation, i.e. number representation without radix; Computing devices using combinations of denominational and non-denominational quantity representations, e.g. using difunction pulse trains, STEELE computers, phase computers using residue arithmetic
- G06F7/728—Methods or arrangements for performing computations using a digital non-denominational number representation, i.e. number representation without radix; Computing devices using combinations of denominational and non-denominational quantity representations, e.g. using difunction pulse trains, STEELE computers, phase computers using residue arithmetic using Montgomery reduction
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Abstract
Description
P·2.sup.n =A·B+Q·N
P·I·2.sup.n ≡N; (Remember that I·2.sup.n ≡1 mod N)
A·B·I+Q·N·I≡AB·I mod N; (Remember that Q·N·I≡0 mod N)
P≡A·B·I mod N.
P≡A·B·I mod N≡(A·B)N.
(P·H)N≡A·mod N;
(P·H)N≡(A·B·I)(H)(I) mod N;
H=I.sup.-2 mod N≡2.sup.2n mod N
P¥(a·B)N,
P=(A·B)N.!
Y=A·B·J mod 2.sup.n (using only the n LS bits);
Z=A·B+(A·B·J mod 2.sup.n)·N.
A·B+(A·B·J mod 2.sup.n)·N! mod 2.sup.n ≡0
J=-N.sup.-1 mod 2.sup.n.
J.sub.0 ≡-N.sub.0.sup.-1 mod 2.sup.k (J.sub.0 exists as N is odd).
S(i-1)<N; B<N and A.sub.i-1 <2.sup.k.
S (i)=Z/2.sup.k (The value of S at the end of the process, before a possible subtraction)
Z=S(i-1)+A.sub.i-1 ·B+(X.sub.0 ·J.sub.0 mod 2.sup.k)N
Z<(N-1)+(2.sup.k -1)·(N-1)+(2.sup.k -1)·N=2.sup.k N+2.sup.k N-N-2.sup.k
Z<2.sup.k ·N+2.sup.k ·N.
Z/2.sup.k <N+N,
______________________________________ Step 1 X = S(0) + A.sub.0 · B = 0 + b · 5c3 = 3f61 Y.sub.0 = X.sub.0 · J.sub.0 mod 2.sup.k = 7 Z = X + Y.sub.0 · N = 3f61 + 7 · a59 = 87d0 S(1) = Z/2.sup.k = 87d (which is smaller than N) Step 2 X = S(1) + A.sub.1 · B =87d + 9 · 5c3 = 3c58 Y.sub.0 = X.sub.0 · J.sub.0 mod 2.sup.k = 8 · 7 mod 2.sup.4 = 8 Z = X + Y.sub.0 · N = 3c58 + 52c8 = 8f20 S(2) = Z/2.sup.k = 8f2 (which is smaller than N) Step 3 X = S(2) + A.sub.2 · B = 8f2 + 9 · 5c3 = 3ccd Y.sub.0 = d · 7 mod 2.sup.4 = b Z = X + Y.sub.0 · N = 3ccd + b · a59 = aea0 S(3) = Z/2.sup.k = aea, as S(3) > N, S(3) = aea - a59 = 91 Therefore C = (A · B)N = 91.sub.16. ______________________________________
______________________________________ Step 1 X = S(0) + C.sub.0 · H = 0 + 1 · 44b = 44b Y.sub.0 = d Z = X + Y.sub.0 · N = 44b + 8685 = 8ad0 S(1) = Z/2.sup.k = 8ad Step 2 X = S(1) + C.sub.1 · H = 8ad + 9 · 44b = 2f50 Y.sub.0 = 0 Z = X + Y.sub.0 · N = 2f50 + 0 = 2f50 S(2) = Z/2.sup.k = 2f5 Step 3 X = S(2) + C.sub.2 · H = 2f5 + 0 · 44b = 2f5 Y.sub.0 = 3 Z = X + Y.sub.0 · N = 2f5 + 3 · a59 = 2200 S(3) = Z/2.sup.k = 220.sub.16 ______________________________________
(A·I.sup.-1)N=A·I.sup.-1 ·I mod N≡A mod N.
______________________________________ Step 1 X = S(0) + A.sub.0 · B = 0 + 9b · 141d =c 2d 8f Y.sub.0 = X.sub.0 · J.sub.0 mod 2.sup.k = 8f · e5 mod 2.sup.8 = eb Z = X + Y.sub.0 · N =c 2d 8f + eb ·2b 13 = 33 b8 00 S(1) ¥ Z/2.sup.k mod N = 33 b8 which is larger than N S(1) = 33 b8 -2b 13 = 8 a5 Step 2 X = S(1) + A.sub.1 · B = 8 a5 + f5 · 141d = 13 48 66 Y.sub.0 = X.sub.0 · J.sub.0 mod 2.sup.k = 66 · E5 mod 2.sup.8 = 3e Z = X + Y.sub.0 · N = 13 48 66 + 3e ·2b 13 = 1d b7 00 S(2) = Z/2.sup.k mod N = 1d b7 Step 3 X = S(2) + A.sub.2 · B = 1d b7 + 0A · 14 1d = e6 d0 Y.sub.0 = d9 · e5 mod 2.sup.8 = 1d Z = X + Y.sub.0 · N = e6 d9 + 1d ·2b 13 = 5 c8 00 S(3) = Z/2.sup.k mod N = 5 c8 ______________________________________
C=A.sup.E mod N.
T=(2.sup.n).sup.Σ mod N=(I.sup.-1).sup.Σ mod N.
______________________________________ Initially: B = A FOR j = 2 TO q B ¥(B · B)N IF E(j) = 1 THEN B ¥(B · A)N END FOR B ¥(B · T)N ______________________________________
E(1)=1; E(2)=0; and E(3)=1,
T=(2.sup.n).sup.Σ mod N=(I.sup.-1 1).sup.Σ mod N.
Σ=2.sup.q-1 +E mod 2.sup.q-1 =2.sup.3-1 +5 mod 2.sup.3-1 =4+1=5
T=14 mod N.
______________________________________ Initially: B = A j = 2, E(2) = 0 B ≡ (B · B)N ≡ A.sup.2 · I mod N j = 3, E(3) = 1 B ≡ (B · B)N ≡ B.sup.2 ≡ A.sup.4 · I.sup.2 · I mod N B ≡ (B · B)N ≡ A.sup.4 · I.sup.3 · A · I mod N and finally: B ¥(B · T)N ≡ A.sup.5 · I.sup.4 · I.sup.-5 · I mod N ≡ A.sup.5 mod ______________________________________ N
C=A.sup.E mod N.
______________________________________ A* ¥(A · H)N B = A* FOR j = 2 TO q-1 B ¥(B · B)N IF E(j) = 1 THEN B ¥(B · A*)N ENDFOR B ¥(B · A)N C = B ______________________________________
B¥(B·1)N instead of B¥(B·A)N
E=1011→E(1)=1; E(2)=0; E(3)=1; E(4)=1;
______________________________________ To find A.sup.1011 mod N; q = 4 A* = (A · H)N = AI.sup.-2 I = AI.sup.-1 mod N B = A* for j = 2 to q B = (B · B)N which produces: A.sup.2 (I.sup.-1).sup.2 · I = A.sup.2 · I.sup.-1 E(2) = 0; B = A.sup.2 · I.sup.-1 j = 3 B = (B · B)N = A.sup.2 (I.sup.-1).sup.2 · I = A.sup.4 · I.sup.-1 E(3) = 1 B = (B · A*)N = (A.sup.4 · I.sup.-1)(AI.sup.-1 ) · I = A.sup.5 · I.sup.-1 j = 4 B = (B · B)N = A.sup.10 · I.sup.-2 · I = A.sup.10 · I.sup.-1 ______________________________________
B=(B·A)=A.sup.10 ·I.sup.-1 ·A·I=A.sup.11
C=B
H=2.sup.2n mod N.
______________________________________ 1 0111 1011 1 0000 0000 1011 SUCCESSFUL SUBTRACT 0101 0 result of the1st round 101 1 NO SUBTRACT 101 00 result of the2nd round 10 11 SUCCESSFUL SUBTRACT 10 010 result of the3rd round 1 011 SUCCESSFUL SUBTRACT 0 1110 result of the4th round 0111 SUCCESSFUL SUBTRACT RESULT 5'TH (n + 1) round 0011 = H (3base 10 = the remainder) ______________________________________
B·A mod N 1)
B.sup.2 mod N, 2)
DO=(DI+CI+XI·Y)mod 2,
CO=CI·XI·Y+CI·DI+DI·Y·XI.
Claims (16)
Priority Applications (1)
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US08/579,951 US5742530A (en) | 1992-11-30 | 1995-12-28 | Compact microelectronic device for performing modular multiplication and exponentiation over large numbers |
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IL10392192 | 1992-11-30 | ||
IL10475393 | 1993-02-16 | ||
IL103921 | 1993-09-06 | ||
IL10692393A IL106923A (en) | 1993-09-06 | 1993-09-06 | Device for performing modular multiplication |
IL104753 | 1993-09-06 | ||
IL106923 | 1993-09-06 | ||
US08/154,220 US5513133A (en) | 1992-11-30 | 1993-11-18 | Compact microelectronic device for performing modular multiplication and exponentiation over large numbers |
US08/579,951 US5742530A (en) | 1992-11-30 | 1995-12-28 | Compact microelectronic device for performing modular multiplication and exponentiation over large numbers |
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US08/154,220 Expired - Lifetime US5513133A (en) | 1992-11-30 | 1993-11-18 | Compact microelectronic device for performing modular multiplication and exponentiation over large numbers |
US08/579,951 Expired - Lifetime US5742530A (en) | 1992-11-30 | 1995-12-28 | Compact microelectronic device for performing modular multiplication and exponentiation over large numbers |
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EP (1) | EP0601907B1 (en) |
JP (1) | JP3636740B2 (en) |
AT (1) | ATE199189T1 (en) |
DE (1) | DE69329929T2 (en) |
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Also Published As
Publication number | Publication date |
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US5513133A (en) | 1996-04-30 |
JP3636740B2 (en) | 2005-04-06 |
DE69329929T2 (en) | 2001-09-27 |
EP0601907A2 (en) | 1994-06-15 |
EP0601907A3 (en) | 1994-11-23 |
EP0601907B1 (en) | 2001-02-14 |
DE69329929D1 (en) | 2001-03-22 |
ATE199189T1 (en) | 2001-02-15 |
JPH07253949A (en) | 1995-10-03 |
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